Show that gravity is described by this 1-form From Harold Edwards' Advanced Calculus: A Differential Forms Approach, section 2.1, exercise 1:

The central force field.  Newton's law of gravitational attraction states that the force exerted by a massive body (the Sun) fixed at the origin $(0,0,0)$ on a particle in space is directed toward the Sun and has magnitude proportional to the inverse square of the distance to the Sun.  Show that this means that the force field is described by the $1$-form $$\frac{kx}{r^3}dx + \frac{ky}{r^3}dy + \frac{kz}{r^3}dz$$ where $k$ is a positive constant and $r=r(x,y,z) = \sqrt{x^2+y^2+z^2}$.

Here's what I've done:
I know that work (at least for small displacements) in physics is defined as $W = F_xdx + F_ydy + F_zdz$ where the force is $\vec F = (F_x, F_y, F_z)$.
I consider the gravity at a point $(x,y,z)$.  Then first the direction toward the Sun is $-x\hat x -y\hat y -z\hat z$.  So the gravity at this point must be proportional to this.
Then I think about the magnitude of $\vec F$ and I get $$\| \vec F\| \propto \frac 1{r^2} \implies {F_x}^2 + {F_y}^2 + {F_z}^2 \propto \frac{1}{r^4}$$
This leads me to think that each component $F_i$ is proportional to $\frac 1{r^2}$.
So putting that together I get that $\vec F$ should be proportional to $$\left(\frac{-x}{r^2},\frac{-y}{r^2},\frac{-z}{r^2}\right)$$
But this problem states that it should be proportional to $$\left(\frac{x}{r^3},\frac{y}{r^3},\frac{z}{r^3}\right)$$
Where have I gone wrong?
 A: Your mistake comes from the assumption that each component of the force should be proportional to $\dfrac 1 {r^2}$. Clearly, this cannot be, since then how would you tell the difference between $F_x$, $F_y$ and $F_z$? Obviously, each of these must somehow contain a $x$, a $y$ and a $z$ respectively. This suggests $F_x \propto \dfrac x {r^3}$ etc. Can you continue it from here?
Edit:
Since the force is directed from the particle to the Sun, it means that, as a vector, it is proportional to the vector that points from the particle to the Sun; since this vector is $-(x,y,z)$, if $f(x,y,z)$ is the proportionality factor, then $\vec F(x,y,z) = -f(x,y,z) (x,y,z)$. Now, if $\| \vec F \| \propto \dfrac 1 {r^2}$, then $| f(x,y,z) | \| (x,y,z) \| \propto \dfrac 1 {r^2}$, and since $\| (x,y,z) \| = r$, we get $|f(x,y,z)| \propto \dfrac 1 {r^3}$, whence $\vec F \propto \dfrac 1 {r^3} (x,y,z)$, where the minus sign has been absorbed into "$\propto$".
To formulate facts about vectors in the language of differential forms, simply replace the vector $(a,b,c)$ with the form $a \Bbb d x + b \Bbb d y + c \Bbb d z$. Note that this has nothing to do with the formula for the work!
